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Inequality

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality $$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$

I stumbled upon this question some days ago and been trying AM-GM to find the solution but so far have been unsuccessful.

Michael Rozenberg
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Kalpan
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    would you mind an answer using functions and derivatives? Or are you looking for something purely elementary? Been a while since I did this stuff... – ShakesBeer Aug 09 '14 at 11:29
  • I have been looking for the elementary approach but your way of solution would be appreciated too. Thanks – Kalpan Aug 09 '14 at 11:34
  • Are you familiar with Lagrangian multipliers? Makes it a lot simpler. – ShakesBeer Aug 09 '14 at 11:36
  • Yes!! I understand the notion of that but still your solution would be appreciated. – Kalpan Aug 09 '14 at 11:43
  • http://en.wikipedia.org/wiki/Lagrange_multiplier here's something to look at if you want while I'm writing (hopefully this will work) – ShakesBeer Aug 09 '14 at 11:46

3 Answers3

4

Ok, here we go.
Let $f(a,b,c,\lambda)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a^3+b^3+c^3}{3}+\lambda(a+b+c-3)$.
$\nabla f=0$ means:

  1. $\frac{\partial f}{\partial a}=-a^2-\frac{1}{a^2}+\lambda=0$
  2. $\frac{\partial f}{\partial b}=-b^2-\frac{1}{b^2}+\lambda=0$
  3. $\frac{\partial f}{\partial c}=-c^2-\frac{1}{c^2}+\lambda=0$

    Now, $ \frac{\partial f}{\partial b}- \frac{\partial f}{\partial a}=0 <=> a^2-b^2+\frac{1}{a^2}-\frac{1}{b^2}=0 <=>$
    $(a-b)(a+b)+(\frac{1}{a}-\frac{1}{b})(\frac{1}{a}+\frac{1}{b})=0 <=>$

    $(a-b)[(a+b)-\frac{1}{ab}(\frac{1}{a}+\frac{1}{b})]=0 <=>$

    $(a-b)(a+b)(1-\frac{1}{(ab)^2})=0$
    And of course the same for all other combinations.
    So $a=b$ or $ab=1$. Also, $b=c$ or $bc=1$.
    Suppose $a=b$. $b=c$ doesn't work (we can test to see it's not a minimum) so we must have $bc=1$.
    Hence we search a value of $a$ with $2a+\frac{1}{a}=3$. This is $a=\frac{1}{2}$. We can test to see this is the minimum (with a value of $\frac{7}{4}$), and we're done.
ShakesBeer
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  • Thanks!! This also explains why the constant in RHS of the inequality can't be 2. – Kalpan Aug 09 '14 at 12:20
  • Glad to help :). Please don't forget to vote up/accept any answers you found useful (accepting is for the answer you found most useful). Welcome to the community. – ShakesBeer Aug 09 '14 at 12:20
0

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that: $$\frac{3v^2}{w^3}\geq\frac{9u^3-9uv^2+w^3}{u^4}+\frac{7}{4u},$$ which is a linear inequality of $v^2$,

which says that it's enough to prove our inequality for an extremal value of $v^2$,

which happens for equality case of two variables.

Let $b=a$.

Hence, $c=3-2a$, where $0<a<\frac{3}{2}$ and we get $$(2a-1)^2(24-45a+26a^2-4a^3)\geq0,$$ which is true for $a\in\left(0,\frac{3}{2}\right)$.

Done!

Michael Rozenberg
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0

here is a elementary method:

WLOG, let $a=$Min{$a,b,c$} $\implies a \le1 ,x=bc$

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=3((a+b+c)^2-3ab-3ac-3bc)=3(9-3a(3-a)-3bc)$

LHS$=\dfrac{a(3-a)+bc}{abc}=\dfrac{a(3-a)+x}{ax},$

RHS$= 9-3a(3-a)-3bc+abc+\dfrac{7}{4}=\dfrac{43}{4}-3a(3-a)+(a-3)x$

LHS-RHS$=\dfrac{(3-a)}{x}\left(x^2-\dfrac{4a^3-36a^2+43a-4}{4a(3-a)}x+1 \right)$

$f(x)=x^2+\dfrac{g_1(a)}{4a(3-a)}x+1 ,g_1(a)=-4a^3+36a^2-43a+4$

it is trivial that $g_1(a)>0 \implies f(x)>0 $

when $g_1(a)<0 ,f_{min}=1-\left(\dfrac{g_1(a)}{2\times4a(3-a)}\right)^2=\dfrac{(2a-1)^2(4-3a)(12a^3-44a^2+67a-4)}{64a^2(3-a)^2}$

$12a^3-44a^2+67a-4+g_1(a)=8a(3-a)>0 \implies 12a^3-44a^2+67a-4>0 \implies \\f_{min} \ge 0$

when $a=\dfrac{1}{2}, f_{min}=0 \implies x= -\dfrac{g_1(a)}{2\times4a(3-a)}=1$

$b+c=\dfrac{5}{2},bc=1 \implies (\dfrac{1}{2},2)$

chenbai
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